Can the Voltage Across a Capacitor Be Greater Than the Source Voltage in a Series RLC Circuit?

  • Thread starter Thread starter Meow12
  • Start date Start date
  • Tags Tags
    Amplitude Voltage
AI Thread Summary
In a series RLC circuit, the voltage across a capacitor can exceed the source voltage due to the relationship defined by the equation V^2 = V_R^2 + (V_L - V_C)^2. The calculations provided yield values for the reactance of the inductor and capacitor, as well as the total impedance and current. The voltage across the resistor, inductor, and capacitor can form a right triangle, allowing for the capacitor's voltage to surpass the source voltage under certain conditions. A higher frequency signal can facilitate this energy storage in the capacitor. The discussion emphasizes the importance of considering phase relationships and complex numbers for a complete understanding of the circuit behavior.
Meow12
Messages
46
Reaction score
20
Homework Statement
You have a ##200\ \Omega## resistor, a ##0.400\ H## inductor, a ##6.00\ \mu F## capacitor, and a voltage source that has a voltage amplitude ##30.0\ V## and an angular frequency of 250 rad/s. They are connected to form an L-R-C series circuit.

(a) What are the voltage amplitudes across the resistor, inductor, and capacitor?
(b) Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.
Relevant Equations
##X_L=\omega L##, ##\displaystyle X_C=\frac{1}{\omega C}##, ##Z=\sqrt{R^2+(X_L-X_C)^2}##
##\displaystyle I=\frac{V}{Z}##
## V_R=IR##, ##V_L=IX_L##, ##V_C=IX_C##
(a) Substituting the values, I get ##X_L=100\ \Omega##, ##X_C=666.67\ \Omega##.

From this, I get ##Z=601\ \Omega##, ##I=49.9\ mA##

##V_R=9.98\ V##, ##V_L=4.99\ V##, ##V_C=33.3\ V##

(b) It's possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source because ##V_R##, ##V_L-V_C##, and ##V## constitute a right triangle where ##V^2=V_R^2+(V_L-V_C)^2##.

-----------------------------------------------------------------------------------------------------------
My answer to (a) matches with the one given in the back of the textbook.
But is my answer to (b) correct?
Thanks.
 
Physics news on Phys.org
Your math checks out. Were they wanting a conceptual answer? If so, something like the frequency of the signal is high enough that it is storing energy in the capacitor.
 
  • Like
Likes topsquark and Meow12
Meow12 said:
(b) It's possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source because VR, VL−VC, and V constitute a right triangle where V2=VR2+(VL−VC)2.

But is my answer to (b) correct?
Yes, but if I was grading, I'd want you to continue your explanation a bit more. It would be customary to express this with complex numbers or magnitudes, since the phase of stuff is critical to understanding it. But that stuff is also pretty clearly implied in your statement.
 
The equation $$ V ^ 2 = V _ R ^ 2 + ( V _ L – V _ C ) ^ 2 $$ is a correct and not complete answer to the question how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source in a series RLC circuit.

The second addend in the equation must be kept constant while the voltage across the inductance and the voltage across the capacitor can be changed, can be increased, so...
 
  • Like
Likes SammyS and Meow12
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...

Similar threads

Back
Top